Vector-Valued Functions

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Objectives:

1.: Understand what a vector valued function is.
2.: Be able to do calculus with vector valued functions: differentiation, integration.
3.: Understand what the derivative of a vector valued function represents geometrically.
4.: Be able to find the arc length of a curve between two points.

Recap Video

You can watch watch the following video which recaps the ideas of the section.

Test your understanding with the following questions.

Parametrization Examples

Below are three videos showing examples of parametrizing curves.

Parametrize the graph in the $xy$ -plane of $x=\sin (y)+y$ .

Parametrize the intersection of the cylinder $x^2+y^2=4$ and the plane $z=1+y$ .

Parametrize the ellipse $x^2+4y^2=9$ .

Problems

Parametrize the intersection of the parabolic sheet $y=x^2$ with the plane $y=1+x+z$ .

If we let $x=t$ , then the parametrization becomes $\textbf {r}(t) = \left \langle t,\answer {t^2},\answer {t^2-t-1} \right \rangle .$

Parametrize the curve in 3D given by $x^2+y^2=1$ in the plane $z=2$ .

We know we can parametrize $x^2+y^2=1$ as $x = \cos (t)$ and $y = \answer {\sin (t)}$ . Therefore, the parametrization for the curve is $\textbf {r}(t) = \left \langle \cos (t),\answer {\sin (t)},\answer {2} \right \rangle , \quad 0\leq t \leq 2\pi .$

Suppose $C$ is the curve given by $\textbf {r}(t) = \left \langle t^2,2t+3,\sin (\pi t) \right \rangle$ .

The vector $\textbf {r}(0) = \left \langle \answer {0},\answer {3},\answer {0} \right \rangle$ .
The derivative $\textbf {r}'(t) = \left \langle \answer {2t},\answer {2},\answer {\pi \cos (\pi t)} \right \rangle$ .
A tangent vector to $C$ at the point $(9,-3,0)$ is $\textbf {r}'(\answer {-3})$ . Plugging this into the derivative gives the vector $\left \langle \answer {-6},\answer {2},\answer {-\pi } \right \rangle .$

Suppose $\textbf {r}'(t) = \left \langle 2t,\sin (t/\pi ),e^{3t}+1 \right \rangle$ . If $\textbf {r}(0) = \left \langle 1,\pi ,1/3 \right \rangle$ , then $\textbf {r}(t) = \left \langle \answer {t^2+1}, \answer {-\pi \cos (t/\pi )+2\pi }, \answer {\frac {1}{3}e^{3t}+t} \right \rangle .$

Suppose a particle is moving with position function $\textbf {r}(t) = \left \langle t^2,1-2t,t \right \rangle$ . At what value of $t$ is the speed of the particle smallest?

Steps:

The velocity of the particle is given by $\textbf {r}'(t) = \left \langle \answer {2t},\answer {-2},\answer {1} \right \rangle$ .
The speed of the particle is $\| \textbf {r} ' (t) \| = \answer {\sqrt {4t^2+5}}$
The speed is smallest at the $t$ -value: $t = \answer {0}$

Find the arc length of the curve $\textbf {r}(t) = \left \langle 2t,\sin (3t),\cos (3t) \right \rangle$ between the points $(0,0,1)$ and $(2\pi ,0,-1)$ .

Steps:

The $t$ -value corresponding to the point $(0,0,1)$ is $t_0 = \answer {0}$
The $t$ -value corresponding to the point $(2\pi ,0,-1)$ is $t_1 = \answer {\pi }$
The velocity vector is $\textbf {r} ' (t) = \left \langle \answer {2},\answer {3\cos (3t)},\answer {-3\sin (3t)} \right \rangle$ .
The speed is $\| \textbf {r} ' (t) \| = \answer {\sqrt {13}}$
The arc length is $\int _{t_0}^{t_1} \|\textbf {r} ' (t)\|dt = \answer {\pi \sqrt {13}} .$

The arc length of the curve $\textbf {r}(t) = \left \langle t\sqrt {2},e^t,e^{-t} \right \rangle$ between the points $(0,1,1)$ and $(\sqrt {2},e,e^{-1})$ is: $\answer {e-e^{-1}}$ .

The expression $2+e^{2t}+e^{-2t}$ factors as $2+e^{2t}+e^{-2t} = (e^t+e^{-t})^2$ .